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How do you prove that $\int_{0}^{1}{\frac{\exp(-rx)}{\sqrt{1-x}}dx}$ converges for $r\geq{1}$ ?

N.B. : I forgot everything about improper integrals, so please be very explicit :)

user63008
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    It actually converges for all $r$. The integrand is less than some constant times $1/\sqrt{1-x}$, so that is all you need to worry about, and you can check that by explicit computation. (If I had time to be more explicit, I'd post this as an answer. Maybe someone else will fill in the details.) – Harald Hanche-Olsen Apr 30 '13 at 11:00
  • @HaraldHanche-Olsen Thank you Harald. Don't hesitate to fill in the details and I will choose your answer ^^ – user63008 Apr 30 '13 at 20:51

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Very explicit: The integrand blows up at $x=1$, so you should be looking at $$\lim_{u\to1-}\int_{0}^{u}\frac{\exp(-rx)}{\sqrt{1-x}}dx.$$ The integrand is positive, so either this diverges to $\infty$ or it remains bounded. Let's show it is bounded. On the given interval, $\exp(-rx)\le1$ if $r>0$, while $\exp(-rx)\le e^{-r}$ if $r$ is negative. In either case just say $\exp(-rx)\le K$. So $$\int_0^u\frac{\exp(-rx)}{\sqrt{1-x}}dx\le\int_0^u\frac{K}{\sqrt{1-x}}dx=2K(1-\sqrt{1-u})<2K$$ for all $u<1$, so the integral is indeed bounded.

Harald Hanche-Olsen
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Let $x=1-u^2$. It suffices to show $\int_0^1 \exp(1-u^2)du$ converges. Note that $\exp(1-u^2)$ is controlled by a constant

Ma Ming
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  • Can you please show me why this change of variable works ? Why does it suffice to show that this other integral converges ? – user63008 Apr 30 '13 at 20:51
  • With that substitution, $\sqrt{1-x}=u$ and $dx=-2u,du$, so the $u$s cancel and your left with the integral shown. (Apart from the factor $2$ and the minus sign, but also $x=0$ yields $u=1$ and $x=1$ yields $u=0$, so the limits of the integrals are interchanged, and another sign change puts them back.) – Harald Hanche-Olsen Apr 30 '13 at 21:05
  • @HaraldHanche-Olsen The next step is to show that $\exp{(1-u^2)}$ is bounded, like in your solution, right ? I guess this is what Ma Ming means by "controlled by a constant". – user63008 Apr 30 '13 at 21:30
  • You're right about that. (I don't often use the phrase “controlled by a constant” myself, but some people seem fond of it.) – Harald Hanche-Olsen May 01 '13 at 06:01